Question : Which of the following is a value of $θ$ , when $\cos^{2} θ - 2 + \cos θ = 0$ ?
Option 1: 60°
Option 2: 90°
Option 3: 30°
Option 4: 0°

Question

Question : If $0^{\circ} < θ < 90^{\circ}$ and $2 \sin^{2} θ + 3 \cos θ = 3$ , then the value of $θ$ is:
Option 1: 30°
Option 2: 60°
Option 3: 45°
Option 4: 75°

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Answer 1

Correct Answer: 0°

Solution : Given, $\cos^{2} θ - 2 + \cos θ = 0$
⇒ $\cos^{2} θ + 2 \cos θ - \cos θ - 2 = 0$
⇒ $\cos θ (\cos θ + 2) - 1 (\cos θ + 2) = 0$
⇒ $(\cos θ + 2) (c o s θ - 1) = 0$
⇒ $\cos θ = - 2$ or $\cos θ = 1$
Since the range of $\cos θ$ is –1 < $\cos θ$ < 1
Hence, $\cos θ$ cannot be –2.
So, $\cos θ = 1 \Rightarrow θ = 0 °$
Hence, the correct answer is 0°.

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Question : Which of the following is a value of θ, when cos2⁡θ−2+cos⁡θ=0?Option 1: 60°Option 2: 90°Option 3: 30°Option 4: 0°

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Question : Which of the following is a value of $θ$ , when $\cos^{2} θ - 2 + \cos θ = 0$ ?
Option 1: 60°
Option 2: 90°
Option 3: 30°
Option 4: 0°